Problem 4 The amount of ore in tons in a segment of a mine i

Problem 4. The amount of ore (in tons) in a segment of a mine is assumed to follow a normal distribution with mean 180 and standard deviation 40. Find the probability that the amount of ore is betwe 172 and 190 tons.

Solution

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    172      
x2 = upper bound =    190      
u = mean =    180      
          
s = standard deviation =    40      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.2      
z2 = upper z score = (x2 - u) / s =    0.25      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.420740291      
P(z < z2) =    0.598706326      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.177966035   [ANSWER]